Answer:The slope on the graph is the price, and it is sloping because the price is cheaper.
Step-by-step explanation:
$10 = 1-6 shirts (amount)
$5 = 7+ shirts (amount)
Hi there!
We can use long division to find the other roots of p(x).
We know that x + 1 is a factor, so:
Set up:
Find how many times that the first term in the divisor goes into the first of the dividend. Subtract from like terms.
x² - x - 6
x + 1 | x³ + 0x² - 7x - 6
x³ + x²
0 - x² - 7x
- x² - x
0 - 6x - 6
-6x - 6
0 0
Therefore, x² - x - 6 is another factor. We can factor this further:
Find two numbers that add up to -1 and multiply into -6. We get:
-3, 2
(x - 3)(x + 2)
The entire polynomial factored is:
(x -3)(x + 2)(x + 1)
Answer: 5.44449
step-by-step explanation:
9514 1404 393
Answer:
- y = -1/7x
- y = 2x -7
Step-by-step explanation:
1. The slope of the given line is the x-coefficient, 7. The slope of the perpendicular line is the negative reciprocal of this: -1/7. Since the line goes through the origin, the y-intercept is zero.
y = -1/7x
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2. The first part of this problem is to find the solution to the given system of equations. We can substitute for x to do that:
3(5 -y)-2y = 10
15 -5y = 10 . . . simplify
-3 +y = -2 . . . . divide by -2
y = 1, x = 4 . . . add 3 to find y; subtract y from 5 to find x
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The second part of this problem requires we note the slope of the parallel line. It is the x-coefficient, 2. The y-intercept can be found from ...
b = y -mx = 1 -2(4) = -7
So, the desired line in y=mx+b form is ...
y = 2x -7
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The first attachment shows the perpendicular lines of problem 1.
The second attachment shows the parallel lines of problem 2.
Answer:
<h3>80° has a tangent of 5.6922..</h3>
Step-by-step explanation:
We are given value of tangent trigonometric ratio = 5.6922.
We need to find the angle in degrees whose measure has a tangent of 5.6922..
So, we need to find the value of .
On plugging in calculator, we get
80.03°.
Now, we need to round it to the nearest whole degree.
Therefore, 80.03°≈ 80°.
<h3>Therefore, 80° has a tangent of 5.6922..</h3>